Question

A capillary tube of radius \(\mathrm{R}\) is immersed in water and water rises in it to a height \(\mathrm{H}\). Mass of water in the capillary tube is M. If the radius of the tube is doubled. Mass of water that will rise in the capillary tube will now be (A) \(\mathrm{M}\) (B) \(2 \mathrm{M}\) (C) \((\mathrm{M} / 2)\) (D) \(4 \mathrm{M}\)

Short answer

The short answer to the problem is (A) \(\mathrm{M}\). When the radius of the capillary tube is doubled, the mass of the water that will rise in the capillary tube will remain the same as the initial mass M.

Step-by-step solution

Express mass in terms of radius, height, and density

We know that mass is the product of volume and density. The volume of the water in the capillary tube can be written as the product of the cross-sectional area of the tube, which is given by \(πR^2\), and the height to which the water rises, which is given by H. Therefore, the volume of the water in the capillary tube can be written as: Volume = \(πR^2H\) Since mass = volume × density, the mass of the water in the tube can be written as: M = ρ (density of water) × \(πR^2H\)

Find the new height when the radius is doubled

Let's assume the new radius is \(2R\). In the problem statement, it's not explicitly mentioned how the height will change when the radius is doubled, but since the same amount of water will be in the new tube, we can relate the heights of the two tubes by saying: \(Volume_1 = Volume_2\) or \(πR^2H = (π(2R)^2H_2)\) Solving this equation for \(H_2\), we get: \(H_2 = \frac{H}{4}\)

Determine the new mass when the radius is doubled

Now that we have the new height, we can find the new mass of the water in the capillary tube: M' (New mass) = ρ (density of water) × \(π(2R)^2H_2\) Substitute the value of \(H_2\) in the equation: M' = ρ (density of water) × \(π(2R)^2\left(\frac{H}{4}\right)\) M' = ρ (density of water) × \(\frac{π(4R^2)H}{4}\) Now, we can rewrite this equation as: M' = ρ (density of water) × \(πR^2H\) Since we already know that: M = ρ (density of water) × \(πR^2H\) Therefore, after canceling out the terms, we can see that: M' = M So, the answer is (A) \(\mathrm{M}\). When the radius of the capillary tube is doubled, the mass of the water that will rise in the capillary tube will remain the same as the initial mass M.

Key concepts

Capillary Tube

A capillary tube is a small, thin, often glass tube with a narrow interior diameter, also known as the radius. It exploits the phenomenon of capillary action, which is the ability of a liquid to flow in narrow spaces without the assistance of external forces like gravity. This is a common occurrence in actions such as the soaking of a paper towel in water or the way plants absorb water from the soil.

In our given scenario, we use the capillary tube to observe how water can rise within this narrow space due to the interplay of cohesive and adhesive forces. Cohesive forces cause the water molecules to stick together, while adhesive forces between the water molecules and the wall of the tube cause the liquid to climb upwards.

Understanding capillary tubes is crucial because they provide insight into how liquids behave on a small scale, which can be applied in various fields, such as biology, chemistry, and physics.

Mass of Water

The mass of water rising in the capillary tube is an important aspect of the problem. The mass ( M) is directly related to the volume and density of the water within the tube. Since we are considering water, a known substance, its density ( ρ) is a constant value provided under stable conditions.

To calculate the mass, we use the formula:

• Mass = Volume × Density

The volume of water in the capillary tube can be derived by considering the formula for the volume of a cylinder:

• Volume = Cross-sectional area × Height

Given the cross-sectional area is πR² (where R is the radius of the tube), and H is the height, the equation becomes:

• Volume = πR²H
• Mass = ρ × πR²H

Here, it becomes evident how the mass is influenced by the dimensions of the capillary tube, specifically its radius and the height to which water rises.

Radius and Height Relationship

The radius and height of the water column within the capillary tube are inversely related due to the principles of volume conservation. If the radius of the capillary tube increases, the height to which the water rises must decrease if the total volume remains the same.

In this problem, when the radius of the tube is doubled to 2R, the height of the water, H₂, changes according to our conservation equation: πR²H = π(2R)²H₂. Solving for H₂, we find that:

• H₂ = H/4

This shows that as the radius increases, the height decreases, maintaining a constant volume of water. Such relationships are essential to comprehend as they assist in predicting how changes in one parameter affect another in physical systems like these tubes.

Volume and Density

Volume and density are critical concepts for understanding the mass of water in a capillary tube. Let's define them clearly:

• Volume refers to the three-dimensional space occupied by a liquid, in this context, the water in the tube.
• Density is the mass per unit volume of the liquid, which for water is generally considered approximately 1 g/cm³ under normal conditions.

These factors combined affect how much water, by mass, the tube can hold.

In our setup, the volume is determined by the geometry of the capillary tube: Volume = πR²H. When considering density, the mass becomes: Mass = ρ × πR²H.

With density as a constant, only changes in volume will influence the mass. It's these principles that help us understand why the mass remains constant even when the tube's radius is altered, as long as it's offset by a corresponding height adjustment.